Question: Simplify $$\sqrt{6+4\sqrt2}+\sqrt{6-4\sqrt2}.$$
Since $(\sqrt2\pm1)^2=2\pm2\sqrt2+1=3\pm2\sqrt2$, $$\sqrt{6+4\sqrt2}=\sqrt{2(3+2\sqrt2)}=\sqrt2(\sqrt2+1)=2+\sqrt2.$$Similarly, $$\sqrt{6-4\sqrt2}=\sqrt2(\sqrt2-1)=2-\sqrt2.$$Therefore, $$\sqrt{6+4\sqrt2}+\sqrt{6-4\sqrt2}=(2+\sqrt2)+(2-\sqrt2)=\boxed{4}.$$